Calculate pH of Buffer Solution from Acetic Acid and NaOH

To find the $\text{pH}$ of the solution $(\text{X})$, we can break down the problem into the following steps:

Step 1: Calculate the amount of acetic acid ($\text{CH}_3\text{COOH}$) present

At complete neutralization of $20\text{ mL}$ of acetic acid with $\text{NaOH}$: $\text{Millimoles of CH}_3\text{COOH} = \text{Millimoles of NaOH used for complete neutralization}$

$\text{Millimoles of CH}_3\text{COOH} = 28.4\text{ mL} \times 0.1\text{ M} = 2.84\text{ mmol}$

Thus, $20\text{ mL}$ of the acetic acid solution contains $2.84\text{ mmol}$ of $\text{CH}_3\text{COOH}$.

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Step 2: Determine the reaction stoichiometry in Solution (X)

Solution $(\text{X})$ is prepared by mixing $20\text{ mL}$ of the acetic acid solution ($2.84\text{ mmol}$) with $14.2\text{ mL}$ of $0.1\text{ M NaOH}$.

$\text{Millimoles of NaOH added} = 14.2\text{ mL} \times 0.1\text{ M} = 1.42\text{ mmol}$

The neutralization reaction is: $\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$

$$\begin{array}{lcccc} & \text{CH}_3\text{COOH} & + & \text{NaOH} & \rightarrow & \text{CH}_3\text{COONa} & + & \text{H}_2\text{O} \\ \text{Initial (mmol):} & 2.84 & & 1.42 & & 0 & & - \\ \text{Final (mmol):} & 1.42 & & 0 & & 1.42 & & - \end{array}$$

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Step 3: Calculate the pH of Solution (X)

After the reaction, the solution contains a mixture of a weak acid ($\text{CH}_3\text{COOH}$) and its conjugate base ($\text{CH}_3\text{COO}^-$ from $\text{CH}_3\text{COONa}$), forming an acidic buffer solution.

Using the Henderson-Hasselbalch equation: $\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{CH}_3\text{COONa}]}{[\text{CH}_3\text{COOH}]}\right)$

Since both the salt and the acid have the same number of millimoles ($1.42\text{ mmol}$) in the same total volume: $\text{pH} = \text{p}K_a + \log_{10}\left(\frac{1.42}{1.42}\right) = \text{p}K_a + \log_{10}(1)$

Since $\log_{10}(1) = 0$: $\text{pH} = \text{p}K_a = 4.75$

Correct Answer: B (4.75)